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Question
at a nature preserve, a wildlife biologist counted bald eagles from an observation deck at the same time each day for 21 days. the table summarizes the resulting data set, data set a. the data value 19 was recorded in error and is removed from data set a to create data set b, which consists of the remaining 20 data values. which statement best compares the median of data set a and the median of data set b?
a. the median of data set b is less than the median of data set a
b. the median of data set b is equal to the median of data set a
c. the median of data set b is greater than the median of data set a
d. there is not enough information to compare the medians of the two data sets
Step1: Find the total number of data points in data set A
The total number of days in data set A is \(1 + 3+4 + 5+4+3 + 1=21\) (using the sum of the number of days row). Since \(n = 21\) (odd), the median is the \(\frac{n + 1}{2}=\frac{21+1}{2}=11^{th}\) ordered - value.
To find the \(11^{th}\) value:
The cumulative frequencies:
For \(0\) bald eagles: \(1\) day
For \(1\) bald eagle: \(1 + 3=4\) days
For \(2\) bald eagles: \(4+4 = 8\) days
For \(3\) bald eagles: \(8 + 5=13\) days. So the median of data set A is \(3\).
Step2: Remove the incorrect data point from data set A to get data set B
Data set A has \(21\) data points. After removing the incorrect data point (value \(19\)), data set B has \(n=20\) data points.
The cumulative frequencies for data set B (excluding the \(19\)):
The new cumulative frequencies:
For \(0\) bald eagles: \(1\) day
For \(1\) bald eagle: \(1+3 = 4\) days
For \(2\) bald eagles: \(4 + 4=8\) days
For \(3\) bald eagles: \(8+5 = 13\) days
For \(4\) bald eagles: \(13+4 = 17\) days
For \(5\) bald eagles: \(17+3=20\) days. Since \(n = 20\) (even), the median is the average of the \(\frac{n}{2}=10^{th}\) and \((\frac{n}{2}+1)=11^{th}\) ordered - values. Both the \(10^{th}\) and \(11^{th}\) values are \(3\). So the median of data set B is \(3\).
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B. the median of data set B is equal to the median of data set A